$\begingroup$The two claims that you've written are logically equivalent to each other. If you prove any one, you've automatically proved the other. A statement of the form $P \implies Q$ is equivalent to the statement $\neg Q \implies \neg P$ (where $\neg$ implies negation, or "not"). The latter is statement is called the "contrapositive" (of the former). So, if you prove that $\alpha \ne \{0\} \implies \alpha = \mathbb R$, then of course, if $\alpha \ne \mathbb R$, then it must be that $\alpha = \{0\}$.$\endgroup$
– M. VinayMar 21 at 1:49

$\begingroup$Assume that $\alpha\neq \mathbb{R}$. If $a\in\alpha$ and $a\neq0$ then there is $a^{-1}\in\mathbb{R}$ such that $a^{-1}a=1$. Since $\alpha$ is ideal, then for all $r\in\mathbb{R}$ you hav ethat $r=r1=(ra^{-1})a\in\alpha$. Contradiction.$\endgroup$
– user647486Mar 21 at 1:50

$\begingroup$Your proof of the first bullet is correct, well done! Let me suggest a simplification though: You don't need to use both 1 and 2 in that step. From 1 itself, you know that $ab = c \in \alpha$, and since $b \ne 0$, you get $a = \frac c b = \frac 1 b \times c \in \alpha$.$\endgroup$
– M. VinayMar 21 at 1:55

2 Answers
2

For the first bullet, I think you could simplify by noting that since $b \in \alpha$ with $b\neq 0$, we have
$$1 = \frac{1}{b} \cdot b \in \alpha$$
as $\frac{1}{b} \in \mathbb{R}$. This implies $\alpha = \mathbb{R}$ by (1).